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Standard XII

Physics

Impulsive Forces in Collision

A particle of...

Question

A particle of mass m moving with velocity u makes an elastic one dimensional collision with a stationary particle of mass m. They are in contact for a short time T. Their force of interaction increases from zero to $F_{0}$ linearly in time $\frac{T}{2}$ and decreases linearly to zero in further time $\frac{T}{2}$ (shown in figure).
The magnitude of $F_{0}$ is :

m u / 2 T

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m u / T

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2 m u / T

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None of these

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Solution

The correct option is B $2 m u / T$
In an elastic one-dimensional collision, particle rebounds with the same speed in opposite direction.
So, change in momentum = 2 mu
But, impulse = $F \times T$ = Change in momentum
$\Rightarrow F_{0} \times T = 2 m u$
or $F_{0} = \frac{2 m u}{T}$

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The correct option is B 2mu/TIn an elastic one-dimensional collision, particle rebounds with the same speed in opposite direction.So, change in momentum = 2 muBut, impulse = F×T = Change in momentum⇒F0×T=2muor F0=2muT

The correct option is B $2 m u / T$
In an elastic one-dimensional collision, particle rebounds with the same speed in opposite direction.
So, change in momentum = 2 mu
But, impulse = $F \times T$ = Change in momentum
$\Rightarrow F_{0} \times T = 2 m u$
or $F_{0} = \frac{2 m u}{T}$